Simplify and expand the following expression: $ \dfrac{4}{3x + 27}- \dfrac{2}{5x + 20}+ \dfrac{4}{x^2 + 13x + 36} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{4}{3x + 27} = \dfrac{4}{3(x + 9)}$ We can factor a $5$ out of denominator in the second term: $ \dfrac{2}{5x + 20} = \dfrac{2}{5(x + 4)}$ We can factor the quadratic in the third term: $ \dfrac{4}{x^2 + 13x + 36} = \dfrac{4}{(x + 9)(x + 4)}$ Now we have: $ \dfrac{4}{3(x + 9)}- \dfrac{2}{5(x + 4)}+ \dfrac{4}{(x + 9)(x + 4)} $ The least common multiple of the denominators is: $ 15(x + 9)(x + 4)$ In order to get the first term over $15(x + 9)(x + 4)$ , multiply by $\dfrac{5(x + 4)}{5(x + 4)}$ $ \dfrac{4}{3(x + 9)} \times \dfrac{5(x + 4)}{5(x + 4)} = \dfrac{20(x + 4)}{15(x + 9)(x + 4)} $ In order to get the second term over $15(x + 9)(x + 4)$ , multiply by $\dfrac{3(x + 9)}{3(x + 9)}$ $ \dfrac{2}{5(x + 4)} \times \dfrac{3(x + 9)}{3(x + 9)} = \dfrac{6(x + 9)}{15(x + 9)(x + 4)} $ In order to get the third term over $15(x + 9)(x + 4)$ , multiply by $\dfrac{15}{15}$ $ \dfrac{4}{(x + 9)(x + 4)} \times \dfrac{15}{15} = \dfrac{60}{15(x + 9)(x + 4)} $ Now we have: $ \dfrac{20(x + 4)}{15(x + 9)(x + 4)} - \dfrac{6(x + 9)}{15(x + 9)(x + 4)} + \dfrac{60}{15(x + 9)(x + 4)} $ $ = \dfrac{ 20(x + 4) - 6(x + 9) + 60} {15(x + 9)(x + 4)} $ Expand: $ = \dfrac{20x + 80 - 6x - 54 + 60}{15x^2 + 195x + 540} $ $ = \dfrac{14x + 86}{15x^2 + 195x + 540}$